Đáp án:
$a) {\left[\begin{aligned}x=6\\x=\frac{-13}{2}\end{aligned}\right.}\\
b)
(x-2y-4)(x-2y+4)\\
c)
x^2(x-1)(x^2+1)\\
d)
{\left[\begin{aligned}x=0\\x=2\end{aligned}\right.}$
$e)x=3$
Giải thích các bước giải:
$a) 2x^2-72+x-6=0\\
\Leftrightarrow 2x^2+x-78=0\\
\Leftrightarrow 2x^2+13x-12x-78=0\\
\Leftrightarrow x(2x+13)-6(2x+13)=0\\
\Leftrightarrow (x-6)(2x+13)=0\\
\Leftrightarrow {\left[\begin{aligned}x-6=0\\2x+13=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=6\\x=\frac{-13}{2}\end{aligned}\right.}\\
b)
x^2-16-4xy+4y^2\\
=x^2-4xy+(2y)^2-4^2\\
=(x-2y)^2-4^2\\
=(x-2y-4)(x-2y+4)\\
c)
x^5-x^4+x^3-x^2\\
=x^4(x-1)+x^2(x-1)\\
=x^2(x-1)(x^2+1)\\
d)
x^4+10x^2-2x^3-20x=0\\
\Leftrightarrow x^2(x^2+10)-2x(x^2+10)=0\\
\Leftrightarrow x(x^2+10)(x-2)=0\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x^2+10=0 (VL)\\ x-2=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x=2\end{aligned}\right.}$
Vì $x^2>0\Rightarrow x^2+10>0$
$e) x^3-9x^2+27x-27=0\\
\Leftrightarrow x^3-3x^2.3+3.x.3^3-3^3=0\\
\Leftrightarrow (x-3)^3=0\\
\Leftrightarrow x-3=0\\
\Leftrightarrow x=3$
